A Proof of Some Important Properties of the Bilinear Form

We demonstrate in this appendix the properties of the bilinear form associated with the geometric
realisation of a Coxeter group . Recall that the matrix

has
only 1’s on the diagonal and non-positive numbers off the diagonal. Recall also that a vector is said to
be positive if and only if all its components are strictly positive, ; this is denoted .
Similarly, a vector is non-negative, , if and only if all its components are non-negative,
. Finally, a vector is non-zero if and only if at least one of its component is non-zero, which is
denoted . Our analysis is based on reference [116]. We shall assume throughout that is
indecomposable.

Main theorem:

The Coxeter group is of finite type if and only if there exists a positive vector
such that .

The Coxeter group is of affine type if and only if there exists a positive vector
such that .

The Coxeter group is of indefinite type if and only if there exists a positive vector
such that .

These cases are mutually exclusive and exhaust all possibilities.

Proof: The proof follows from a series of lemmata. The inequalities define a convex cone, namely
the first quadrant . Similarly, the inequalities define also a convex cone . One has
indeed:

Note that one has also

and . There are three distinct cases for the intersection :

Case 1: .

Case 2: .

Case 3: .

These three distinct cases correspond, as we shall now show, to the three distinct cases of the theorem.
To investigate these distinct cases, we need the following lemmata:

Lemma 1: The conditions and imply either or . In other
words

Proof: Assume that fulfills and has at least one component equal to zero. We shall show
that all its components are then zero. Assume for and for . One
has (with no non-vanishing component if ). From one gets
. Take . As in the previous sum, one has
and thus , which implies . As (), this leads to
for and . The matrix would be decomposable, unless , i.e. when all
components vanish.

Lemma 2: Consider the system of linear homogeneous inequalities

on the vector . This system possesses a solution if and only if there is no set of numbers that
are not all zero such that .

Proof: This is a classical result in the theory of linear inequalities (see [116], page 47).

Case 1: ,

by Lemma 1. Furthermore, cannot contain a nontrivial subspace since implies
, but only one of the two can be in when . Hence , i.e.,
and

This excludes in particular the existence of a vector such that or .

Finally, the interior of is non-empty since is nondegenerate. Taking a non-zero vector
such that , one concludes that there exists a vector such that . This shows that
Case 1 corresponds to the first case in the theorem. We shall verify below that is indeed positive
definite.

Case 2: ,

reduces in that case to a straight line. Indeed, let be an element of and let
be in but not in . Let be the straight line joining and . Consider the line
segment from to . This line segment is contained in and crosses the boundary of at
some point . But by Lemma 1, this point must be the origin. Thus, , for some real number
. This implies that the entire line is in since for all , and
also for all since .

Let be any other point in . If , the segment joining to intersects and this
can only be at the origin by Lemma 1. Hence . If , the segment joining to intersects
and this can only be at the origin by Lemma 1. Hence, we find again that . This shows that
reduces to the straight line .

Since , one has . Hence, , which excludes
the existence of a vector such that (one would have and hence ).
Furthermore, there exists such that . This shows that Case 2 corresponds to the second
case in the theorem. We shall verify below that is indeed positive semi-definite. Note that
and that the corank of is one.

Case 3:

In that case, there is a vector such that , which corresponds to the third case in the
theorem. Indeed, consider the system of homogeneous linear inequalities

By Lemma 2, this system possesses a solution if and only if there is no non-trivial such
that .

Consider thus the equations for , or, as is symmetric,
. Since , these conditions are equivalent to (if , one
defines through ), i.e., . But , i.e., , which implies
and hence also . The all vanish and the general solution to the equations
is accordingly trivial.

To conclude the proof of the main theorem, we prove the following proposition:

Proposition: The Coxeter group belongs to Case 1 if and only if is positive definite; it belongs to
Case 2 if and only if is positive semi-definite with .

Proof: If is positive semi-definite, then it belongs to Case 1 or Case 2 since otherwise
there would be a vector such that and thus , leading to a
contradiction. In the finite case, is positive definite and hence, : This corresponds to
Case 1. In the affine case, there are zero eigenvectors and : This corresponds to
Case 2.

Conversely, assume that the Coxeter group belongs to Case 1 or Case 2. Then there exists a vector
such that . This yields for and therefore belongs to
Case 1. In particular, , which shows that the eigenvalues of are
all non-negative: is positive semi-definite. We have seen furthermore that it has the eigenvalue zero
only in Case 2.